Retailer’s inventory decisions with promotional efforts and preservation technology investments when supplier offers quantity discounts

Abstract

In this paper, we formulate a mathematical inventory model for deteriorating items with a constant rate of deterioration. The supplier offers the retailer successive discount on purchase of goods if the order size crosses predefined quantity levels. Retailer uses preservation techniques to reduce the deterioration rate. In order to increase the sales, retailer implements promotional strategies. Market demand of the product is influenced by promotional efforts, stock level and selling price of the product. The objective is to find optimum order quantity, cycle time, promotional cost, preservation cost and selling price in order to maximize total profit for the retailer. A numerical example is given to validate the mathematical model. Sensitivity analysis has been carried out to analyze the effect of change of other inventory parameters on decision variables and total profit. Results indicate that due to preservation technology we can see remarkable decrease in the deterioration rate hence cycle time increases and retailer can set a cheaper selling price to increase sales. Promotional efforts help the retailer to enhance sales of the product. Moreover, depending upon the product demand and order size, quantity discounts help retailer to reduce the purchase cost and hence overall profit of retailer increases.

Appendix

For $0 \le t \le T$, we have.

$\frac{d}{dt}I\left( t \right) + \left( {\theta + b} \right) \cdot I\left( t \right) = - \left( {a - c \cdot p + \beta \cdot M} \right)\,$. This is a linear differential equation.

Integrating Factor, $I.F. = e^{{\left( {\theta + b} \right)t}}$. Hence the $I\left( t \right)$ can be obtained by solving following equation.

$I\left( t \right) \cdot e^{{\left( {\theta + b} \right)t}} = k - \int\limits_{0}^{t} {\left( {a - cp + \beta M} \right)e^{{\left( {\theta + b} \right)t}} dt}$ where, k is an arbitrary constant.

Therefore, $I\left( t \right) \cdot e^{{\left( {\theta + b} \right)t}} = k - \left[ {\frac{{\left( {a - cp + \beta M} \right)}}{{\left( {\theta + b} \right)}}e^{{\left( {\theta + b} \right)t}} } \right]_{0}^{t}$. Simplifying further we have,

$I\left( t \right) \cdot e^{{\left( {\theta + b} \right)t}} = k - \left[ {\frac{{\left( {a - cp + \beta M} \right)}}{{\left( {\theta + b} \right)}}\left\{ {e^{{\left( {\theta + b} \right)t}} - 1} \right\}} \right]$. Now use the boundary condition $I\left( T \right) = 0$.

Hence, $k = \left[ {\frac{{\left( {a - cp + \beta M} \right)}}{{\left( {\theta + b} \right)}}\left\{ {e^{{\left( {\theta + b} \right)T}} - 1} \right\}} \right]$. Substituting this value in above equation we get,

$I\left( t \right) \cdot e^{{\left( {\theta + b} \right)t}} = \frac{{\left( {a - cp + \beta M} \right)}}{{\left( {\theta + b} \right)}}\left\{ {e^{{\left( {\theta + b} \right)T}} - e^{{\left( {\theta + b} \right)t}} } \right\}$. Simplifying further we obtain $I\left( t \right)$ as shown below.

$I\left( t \right) = \frac{{\left( {a - cp + \beta M} \right)}}{{\left( {\theta + b} \right)}}\left\{ {e^{{\left( {\theta + b} \right)\left( {T - t} \right)}} - 1} \right\}$ Hence Eq. (4) is obtained.